Munkres Topology Solution: Chapter 8

§48 Baire Spaces

Ex.48.1

Let $X$ equal the countable union $\bigcup B_n$. Show that if $X$ is a nonempty Baire space, at least one of the sets $\overline{B_n}$ has a nonempty interior.

$X=\bigcup\overline{B_n}$. If every $\overline{B_n}$ has empty interior, since $X$ is Baire, their union has empty interior, contradiction.

Ex.48.2

The Baire category theorem implies that $\Bbb R$ cannot be written as a countable union of closed subsets having empty interiors. Show this fails if the sets are not required to be closed.

$\Bbb R=\Bbb Q\cup(\Bbb R\setminus\Bbb Q)$.

Ex.48.3

Show that every locally compact Hausdorff space is a Baire space.

A direct proof goes similarly to compact Hausdorff case. Let $A_n$ be closed sets with empty interiors and $U_0\subset X$ be any open set. Since $\mathrm{int} A_1=\emptyset$, $U_0\not\subset A_1$. So there exists $x\in U_0\setminus A_1$. Since $X$ is regular, we can find open nbhd $U_1$ about $x$ such that $U_0\supset \overline{U_1}\supset U_1$. And, since $X$ is locally compact, we can choose $U_1$ such that it has compact closure. Proceed recursively, we have $U_0\supset \overline{U_1}\supset U_1\supset \overline{U_2}\supset\dots$, with $U_i\not\subset A_{i+1}$ and $\overline{U_i}$ being compact. $\overline{U_1}$ is compact, thus having finite intersection property: $\bigcap U_i\neq\emptyset$. Any $x\in\bigcap U_i$ will lie in $U_0$ but outside of $\bigcup A_i$.

An alternative proof is: the one-point compactification of $X$ is Baire, while $X$ is its open subspace, hence $X$ is Baire.

Ex.48.4

Show that if every point $x$ of $X$ has a neighborhood that is a Baire space, then $X$ is a Baire space. [Hint: Use the open set formulation of the Baire condition.]

Let $U_n$ be a sequence of dense open sets. For each point $x\in X$, it has an open nbhd $U$ that is itself Baire. $U\cap U_n$ is a sequence of dense open sets in $U$, thus $U\cap\bigcap U_n$ is dense in $U$. In particular, $U\cap\bigcap U_n\neq\emptyset$. Now if $V$ is an arbitrary nonempty open set, choose such $U$ above and note that $U\cap V$ is Baire, thus $U\cap V\cap\bigcap U_n\neq\emptyset$.

Ex.48.5

Show that if $Y$ is a $G_\delta$ set in $X$, and if $X$ is compact Hausdorff or complete metric, then $Y$ is a Baire space in the subspace topology. [Hint: Suppose that $Y = \bigcap W_n$, where $W_n$ is open in $X$, and that $B_n$ is closed in $Y$ and has empty interior in $Y$. Given $U_0$ open in $X$ with $U_0\cap Y\neq\emptyset$, find a sequence of open sets $U_n$ of $X$ with $U_n\cap Y$ nonempty, such that
$$
\begin{aligned}
\overline{U_n}&\subset U_{n-1}\\
\overline{U_n}\cap\overline{B_n}&=\emptyset\\
\mathrm{diam}U_n&<1/n, \text{in the metric case}\\
\overline{U_n}&\subset W_n.]
\end{aligned}
$$

Note that topological completeness is $G_\delta$ hereditary, so we don’t need to prove the metric case. For the other case, suppose $Y=\bigcap W_n$. Since compactness and Hausdorff-ness are closed hereditary, we can WLOG assume $Y$ is dense. Suppose $V_n$ is a sequence of dense open sets in $Y$, thus dense in $X$. Assume $V_n=U_n\cap Y$ where $U_n$ are open sets in $X$. Then $\bigcap V_n=\bigcap U_n\cap\bigcap W_n$. However, since $U_n$ and $W_n$ are all dense open sets, we know $\bigcap V_n$ is dense.

Ex.48.6

Show that the irrationals are a Baire space.

$\Bbb R\setminus\Bbb Q$ is a $G_\delta$ subset of $\Bbb R$.

Ex.48.7

Prove the following:
Theorem. If $D$ is a countable dense subset of $\Bbb R$, there is no function $f : \Bbb R\to\Bbb R$ that is continuous precisely at the points of $D$.
Proof.
(a) Show that if $f: \Bbb R\to\Bbb R$, then the set $C$ of points at which $f$ is continuous is a $G_\delta$ set in $\Bbb R$. [Hint: Let $U_n$ be the union of all open sets $U$ of $\Bbb R$ such that $\mathrm{diam} f (U) < 1/n$. Show that $C =\bigcap U_n$.]
(b) Show that $D$ is not a $G_\delta$ set in $\Bbb R$. [Hint: Suppose $D = \bigcap W_n$, where $W_n$ is open in $\Bbb R$. For $d\in D$, set $V_d = \Bbb R − \set{d}$. Show Wn and $V_d$ are dense in $\Bbb R$.]

(a) $$
\set{x|f\text{ continuous at } x}=\bigcap_n\bigcup_m\set{x|\forall s,t\in(x-\frac1m,x+\frac1m),|f(s)-f(t)|<\frac1n},
$$ which is a $G_\delta$ set.

(b) Generally, a $G_\delta$ dense subset of $\Bbb R$ cannot be countable. Otherwise $D=\bigcap W_n$ is an intersection of countable dense open sets. By Baire category theorem, $\bigcap W_n\cap\bigcap _{d\in D}(\Bbb R\setminus\set{d})$ is dense. But this equals $\emptyset$, a contradiction.

Ex.48.8

If $f_n$ is a sequence of continuous functions $f_n : \Bbb R\to \Bbb R$ such that $f_n(x) → f (x)$ for each $x\in\Bbb R$, show that $f$ is continuous at uncountably many points of $\Bbb R$.

By theorem 48.5, the set of continuous point of $f$ is dense. By Ex.48.7, it is not countable.

Ex.48.9

Let $g :\Bbb Z_+\to\Bbb Q$ be a bijective function; let $x_n = g(n)$. Define $f : \Bbb R\to\Bbb R$ as follows: $f(x_n) = 1/n$ for $x_n\in\Bbb Q$, $f(x) =0$ for $x\notin\Bbb Q$. Show that $f$ is continuous at each irrational and discontinuous at each rational. Can you find a sequence of continuous functions fn coverging to $f$?

That $f$ is continuous precisely at irrationals is a standard fact in analysis. However, I haven’t proven or denied the existence of such $f_n$…

Ex.48.10

Prove the following:
Theorem (Uniform boundedness principle). Let $X$ be a complete metric space, and let $\mathcal F$ be a subset of $C(X,\Bbb R)$ such that for each $a\in X$, the set $$\mathcal F_a = \set{ f (a) | f\in\mathcal F }$$ is bounded. Then there is a nonempty open set $U$ of $X$ on which the functions in $\mathcal F$ are uniformly bounded, that is, there is a number $M$ such that $|f (x)|\le M$ for all $x\in U$ and all $f\in\mathcal F$ . [Hint: Let $A_N = \set{x; | f (x)|\le N \text{ for all } f\in F }$.]

We follow the hint. Clearly $X=\bigcup A_n$ since $\mathcal F$ is pointwise bounded, and $A_n$ is a sequence of closed sets. By Ex.48.1, there is at least one set, say $A_{n_0}$, has nonempty interior, and write $U=\mathrm{int}A_{n_0}$. By definition this is what we want.

Ex.48.11

Determine whether or not $\Bbb R_l$ is a Baire space.

Yes. First we claim: a subset $D\subset \Bbb R$ is dense in usual topology, if, and only if it is dense in lower limit topology. Suppose it is dense in usual topology. Now given any point $x$, any open nbhd (in $\Bbb R_l$) $[x,x+\epsilon)$, note that $D\cap [x,\infty)$ is also dense in $[x,\infty)$ in usual topology, and $[x,x+\epsilon)$ is open in this topology. Hence $[x,x+\epsilon)\cap D\neq\emptyset$. (NOTE: I am not asserting that if $D$ dense, $A$ closed, then $D\cap A$ dense in $A$! This is true when $\mathrm{int}A\neq\emptyset$, or when $A$ is open. You can also check this claim directly.)

As for this problem, let $W_n=\bigcup_{\alpha\in J_n}[a_{n,\alpha},b_{n,\alpha})$ be a sequence of dense open sets in $\Bbb R_l$. Now note that any open nbhd (in $\Bbb R_l$) intersects $W_n$, it must intersects some $(a_{n,\alpha},b_{n,\alpha})$. i.e. $W_n’=\bigcup_{\alpha\in J_n}(a_{n,\alpha},b_{n,\alpha})$ is a sequence of dense open sets in $\Bbb R_l$, so dense in $\Bbb R$. Therefore, $\bigcup W_n\supset\bigcup W_n’$ is dense.

Ex.48.12

Show that $\Bbb R^J$ is a Baire space in the box, product, and uniform topologies.

For uniform topology $\Bbb R^J$ is clearly Baire, since it is complete.

For product topology, there is no hope for $\Bbb R^J$ to be locally compact or complete metric, since it is not normal if $|J|\ge \mathfrak c$. But we can mimic the proof of theorem 49.2 again. All we need to prove is: if $U_0\supset\overline{U_1}\supset U_1\supset…$, then $\bigcap U_n\neq\emptyset$. However, this follows easily from the fact that a descending sequence of CLOSED intervals has nonempty intersection.

The same applies to box topology.

Ex.48.13

Let $X$ be a topological space; let $Y$ be a complete metric space. Show that $C(X, Y )$ is a Baire space in the fine topology (see Exercise 11 of §46). [Hint: Given basis elements $B( f_i, \delta_i )$ such that $\delta_1\le 1$ and $\delta_{i+1}\le \delta_i /3$ and $f_{i+1}\in B( f_i, δ_i /3)$, show that $\bigcap B( f_i, \delta_i )\neq \emptyset$.]

Again, we mimic the proof of theorem 49.2.

First, we show $C(X,Y)$ in the fine toplogy is regular. This is easy. Given $f\in B(f,\delta)$, we need to show $\overline{B(f,\delta/2)}\subset B(f,\delta)$. Suppose $g\notin B(f,\delta)$, then $B(g,\delta/2)\cap B(f,\delta/2)=\emptyset$. Done.

Next, follow the proof of theorem 49.2, we get a descending chain: $B(f_1,\delta_1)\supset\overline{B(f_2,\delta_2)}\supset B(f_2,\delta_2)\supset…$ with $\delta_1\le 1$ and $\delta_{i+1}\le \delta_i /3$. We are going to show it has nonempty limit. For any $\epsilon>0$, find $N$ such that $(1/3)^N<\epsilon$, then for $m>n\ge N$, $f_m, f_n\in B(f_N,\delta_N)$, so $\bar\rho(f_n,f_N)<\epsilon, \bar\rho(f_m,f_N)<\epsilon$. Hence $\bar\rho(f_n,f_m)<2\epsilon$. This implies that $f_n$ is a Cauchy sequence in the uniform topology. Note that $C(X,Y)$ in the uniform topology is complete, so it has a uniform limit $f$, which is continuous. Therefore, $f\in\bigcap B(f_n,\delta_n)$.

执笔至此，发现维数理论已经全部忘记，索性就写到这吧。维数理论我应该会在准备博资考复习的时候再看一遍，到时通过考试后应该会补充上的，包括前面一些没写的，比如Hausdorff metric的题目等等。Munkres前面有一题总结所有的基本拓扑性质，我写了一张很大的表，展示了许多基本性质的互推关系和不互推的反例，当然并不能覆盖所有的关系，感觉总结出所有的太枯燥了，也没太大意义，有些反例也非常弯弯绕绕，看一遍就忘掉，就算记住也不知道有甚用。明年如果我会做一般拓扑学助教的话就把常用的找个时间做成一张图放出来。